Content check — Business Statistics, Chapter 8: Confidence Intervals

8-2 §8.2 source ↗

Use the following information to answer the next five exercises. A hospital is trying to cut down on emergency room wait times. It is interested in the amount of time patients must wait before being called back to be examined. An investigation committee randomly surveyed 70 patients. The sample mean was 1.5 hours with a sample standard deviation of 0.5 hours.

Define the random variables X and $\bar{X}$ in words.

Answer

X is the number of hours a patient waits in the emergency room before being called back to be examined.

\bar{X}

is the mean wait time of 70 patients in the emergency room.

8-4 §8.2 source ↗

Use the following information to answer the next five exercises. A hospital is trying to cut down on emergency room wait times. It is interested in the amount of time patients must wait before being called back to be examined. An investigation committee randomly surveyed 70 patients. The sample mean was 1.5 hours with a sample standard deviation of 0.5 hours.

Construct a 95% confidence interval for the population mean time spent waiting. State the confidence interval, sketch the graph, and calculate the error bound.

Answer

CI: (1.3808, 1.6192)

EBM = 0.12

This is a normal distribution curve. The peak of the curve coincides with the point 1.5 on the horizontal axis. A central region is shaded between points 1.38 and 1.62. — Figure 8.12 in the book ↗

8-6 §8.2 source ↗

Use the following information to answer the next six exercises: One hundred eight Americans were surveyed to determine the number of hours they spend watching television each month. It was revealed that they watched an average of 151 hours each month with a standard deviation of 32 hours. Assume that the underlying population distribution is normal.

Identify the following:

a. $\bar{x}$ =_______

b. $s_{x}$ =_______

c. n =_______

d. n – 1 =_______

Answer

a. $\bar{x}$ = 151

b. $s_{x}$ = 32

c. n = 108

d. n – 1 = 107

8-8 §8.2 source ↗

Use the following information to answer the next six exercises: One hundred eight Americans were surveyed to determine the number of hours they spend watching television each month. It was revealed that they watched an average of 151 hours each month with a standard deviation of 32 hours. Assume that the underlying population distribution is normal.

Define the random variable $\bar{X}$ in words.

Answer

\bar{X}

is the mean number of hours spent watching television per month from a sample of 108 Americans.

8-10 §8.2 source ↗

Use the following information to answer the next six exercises: One hundred eight Americans were surveyed to determine the number of hours they spend watching television each month. It was revealed that they watched an average of 151 hours each month with a standard deviation of 32 hours. Assume that the underlying population distribution is normal.

Construct a 99% confidence interval for the population mean hours spent watching television per month. (a) State the confidence interval, (b) sketch the graph, and (c) calculate the error bound.

Answer

CI: (142.92, 159.08)

EBM = 8.08

This is a normal distribution curve. The peak of the curve coincides with the point 151 on the horizontal axis. A central region is shaded between points 142.92 and 159.08. — Figure 8.13 in the book ↗

8-12 §8.2 source ↗

Use the following information to answer the next 13 exercises: The data in Table 8.2 are the result of a random survey of 39 national flags (with replacement between picks) from various countries. We are interested in finding a confidence interval for the true mean number of colors on a national flag. Let X = the number of colors on a national flag.

Calculate the following:

a. $\bar{x}$ =______

b. $s_{x}$ =______

c. n =______

Table 8.2 in the book ↗

X	Freq.
1	1
2	7
3	18
4	7
5	6

Answer

a. 3.26

b. 1.02

c. 39

8-14 §8.2 source ↗

Use the following information to answer the next 13 exercises: The data in Table 8.2 are the result of a random survey of 39 national flags (with replacement between picks) from various countries. We are interested in finding a confidence interval for the true mean number of colors on a national flag. Let X = the number of colors on a national flag.

What is $\bar{x}$ estimating?

Table 8.2 in the book ↗

X	Freq.
1	1
2	7
3	18
4	7
5	6

Answer

μ

8-16 §8.2 source ↗

Use the following information to answer the next 13 exercises: The data in Table 8.2 are the result of a random survey of 39 national flags (with replacement between picks) from various countries. We are interested in finding a confidence interval for the true mean number of colors on a national flag. Let X = the number of colors on a national flag.

As a result of your answer to Exercise 8.15, state the exact distribution to use when calculating the confidence interval.

Table 8.2 in the book ↗

X	Freq.
1	1
2	7
3	18
4	7
5	6

Answer

t ₃₈

8-18 §8.2 source ↗

Construct a 95% confidence interval for the true mean number of colors on national flags.

How much area is in each tail?

Answer

0.025

8-20 §8.2 source ↗

Construct a 95% confidence interval for the true mean number of colors on national flags.

The 95% confidence interval is_____.

Answer

(2.93, 3.59)

8-22 §8.2 source ↗

Construct a 95% confidence interval for the true mean number of colors on national flags.

In one complete sentence, explain what the interval means.

Answer

We are 95% confident that the true mean number of colors for national flags is between 2.93 colors and 3.59 colors.

8-23 §8.2 source ↗

Construct a 95% confidence interval for the true mean number of colors on national flags.

Using the same $\bar{x}$ , $s_{x}$ , and level of confidence, suppose that n were 69 instead of 39. Would the error bound become larger or smaller? How do you know?

Answer

The error bound would become EBM = 0.245. This error bound decreases because as sample sizes increase, variability decreases and we need less interval length to capture the true mean.

8-26 §8.3 source ↗

Use the following information to answer the next two exercises: Marketing companies are interested in knowing the population percent of women who make the majority of household purchasing decisions.

If it were later determined that it was important to be more than 90% confident and a new survey were commissioned, how would it affect the minimum number you need to survey? Why?

Answer

The sample size needed would increase. As the confidence level increases,

α

decreases and

z_{(\frac{a}{2})}

increases. To maintain the same error bound, the size of the sample needs to increase.

8-28 §8.3 source ↗

Use the following information to answer the next five exercises: Suppose the marketing company did do a survey. They randomly surveyed 200 households and found that in 120 of them, the woman made the majority of the purchasing decisions. We are interested in the population proportion of households where women make the majority of the purchasing decisions.

Define the random variables X and $p^{'}$ in words.

Answer

X is the number of “successes” where the woman makes the majority of the purchasing decisions for the household. P′ is the percentage of households sampled where the woman makes the majority of the purchasing decisions for the household.

8-30 §8.3 source ↗

Use the following information to answer the next five exercises: Suppose the marketing company did do a survey. They randomly surveyed 200 households and found that in 120 of them, the woman made the majority of the purchasing decisions. We are interested in the population proportion of households where women make the majority of the purchasing decisions.

Construct a 95% confidence interval for the population proportion of households where the women make the majority of the purchasing decisions. State the confidence interval, sketch the graph, and calculate the error bound.

Answer

CI: (0.5321, 0.6679)

EBM: 0.0679

This is a normal distribution curve. The peak of the curve coincides with the point 0.6 on the horizontal axis. A central region is shaded between points 0.5321 and 0.6679. — Figure 8.14 in the book ↗

8-32 §8.3 source ↗

Use the following information to answer the next five exercises: Of 1,050 randomly selected adults, 360 identified themselves as manual laborers, 280 identified themselves as non-manual wage earners, 250 identified themselves as mid-level managers, and 160 identified themselves as executives. In the survey, 82% of manual laborers preferred trucks, 62% of non-manual wage earners preferred trucks, 54% of mid-level managers preferred trucks, and 26% of executives preferred trucks.

We are interested in finding the 95% confidence interval for the percent of executives who prefer trucks. Define random variables X and P′ in words.

Answer

X is the number of “successes” where an executive prefers a truck. P′ is the percentage of executives sampled who prefer a truck.

8-34 §8.3 source ↗

Use the following information to answer the next five exercises: Of 1,050 randomly selected adults, 360 identified themselves as manual laborers, 280 identified themselves as non-manual wage earners, 250 identified themselves as mid-level managers, and 160 identified themselves as executives. In the survey, 82% of manual laborers preferred trucks, 62% of non-manual wage earners preferred trucks, 54% of mid-level managers preferred trucks, and 26% of executives preferred trucks.

Construct a 95% confidence interval. State the confidence interval, sketch the graph, and calculate the error bound.

Answer

CI: (0.19432, 0.33068)

This is a normal distribution curve. The peak of the curve coincides with the point 0.26 on the horizontal axis. A central region is shaded between points 0.1943 and 0.3307. — Figure 8.15 in the book ↗

8-36 §8.3 source ↗

Use the following information to answer the next five exercises: Of 1,050 randomly selected adults, 360 identified themselves as manual laborers, 280 identified themselves as non-manual wage earners, 250 identified themselves as mid-level managers, and 160 identified themselves as executives. In the survey, 82% of manual laborers preferred trucks, 62% of non-manual wage earners preferred trucks, 54% of mid-level managers preferred trucks, and 26% of executives preferred trucks.

The sampling error given in the survey is ±2%. Explain what the ±2% means.

Answer

The sampling error means that the true mean can be 2% above or below the sample mean.

8-38 §8.3 source ↗

Use the following information to answer the next five exercises: A poll of 1,200 voters asked what the most significant issue was in the upcoming election. Sixty-five percent answered the economy. We are interested in the population proportion of voters who feel the economy is the most important.

Define the random variable P′ in words.

Answer

P′ is the proportion of voters sampled who said the economy is the most important issue in the upcoming election.

8-40 §8.3 source ↗

Use the following information to answer the next five exercises: A poll of 1,200 voters asked what the most significant issue was in the upcoming election. Sixty-five percent answered the economy. We are interested in the population proportion of voters who feel the economy is the most important.

Construct a 90% confidence interval, and state the confidence interval and the error bound.

Answer

CI: (0.62735, 0.67265)

EBM: 0.02265

8-42 §8.3 source ↗

Use the following information to answer the next 16 exercises: The Ice Chalet offers dozens of different beginning ice-skating classes. All of the class names are put into a bucket. The 5 P.M., Monday night, ages 8 to 12, beginning ice-skating class was picked. In that class were 64 girls and 16 boys. Suppose that we are interested in the true proportion of girls, ages 8 to 12, in all beginning ice-skating classes at the Ice Chalet. Assume that the children in the selected class are a random sample of the population.

What is being counted?

Answer

The number of girls, ages 8 to 12, in the 5 P.M. Monday night beginning ice-skating class.

8-44 §8.3 source ↗

Use the following information to answer the next 16 exercises: The Ice Chalet offers dozens of different beginning ice-skating classes. All of the class names are put into a bucket. The 5 P.M., Monday night, ages 8 to 12, beginning ice-skating class was picked. In that class were 64 girls and 16 boys. Suppose that we are interested in the true proportion of girls, ages 8 to 12, in all beginning ice-skating classes at the Ice Chalet. Assume that the children in the selected class are a random sample of the population.

Calculate the following:

a. x = _______

b. n = _______

c. p′ = _______

Answer

a. x = 64

b. n = 80

c. p′ = 0.8

8-46 §8.3 source ↗

Use the following information to answer the next 16 exercises: The Ice Chalet offers dozens of different beginning ice-skating classes. All of the class names are put into a bucket. The 5 P.M., Monday night, ages 8 to 12, beginning ice-skating class was picked. In that class were 64 girls and 16 boys. Suppose that we are interested in the true proportion of girls, ages 8 to 12, in all beginning ice-skating classes at the Ice Chalet. Assume that the children in the selected class are a random sample of the population.

Define a new random variable P′. What is p′ estimating?

Answer

p

8-48 §8.3 source ↗

Use the following information to answer the next 16 exercises: The Ice Chalet offers dozens of different beginning ice-skating classes. All of the class names are put into a bucket. The 5 P.M., Monday night, ages 8 to 12, beginning ice-skating class was picked. In that class were 64 girls and 16 boys. Suppose that we are interested in the true proportion of girls, ages 8 to 12, in all beginning ice-skating classes at the Ice Chalet. Assume that the children in the selected class are a random sample of the population.

State the estimated distribution of P′. Construct a 92% Confidence Interval for the true proportion of girls in the ages 8 to 12 beginning ice-skating classes at the Ice Chalet.

Answer

P^{'} ~ N (0.8, \sqrt{\frac{(0.8) (0.2)}{80}})

. (0.72171, 0.87829).

8-50 §8.3 source ↗

Use the following information to answer the next 16 exercises: The Ice Chalet offers dozens of different beginning ice-skating classes. All of the class names are put into a bucket. The 5 P.M., Monday night, ages 8 to 12, beginning ice-skating class was picked. In that class were 64 girls and 16 boys. Suppose that we are interested in the true proportion of girls, ages 8 to 12, in all beginning ice-skating classes at the Ice Chalet. Assume that the children in the selected class are a random sample of the population.

How much area is in each tail?

Answer

0.04

8-52 §8.3 source ↗

Use the following information to answer the next 16 exercises: The Ice Chalet offers dozens of different beginning ice-skating classes. All of the class names are put into a bucket. The 5 P.M., Monday night, ages 8 to 12, beginning ice-skating class was picked. In that class were 64 girls and 16 boys. Suppose that we are interested in the true proportion of girls, ages 8 to 12, in all beginning ice-skating classes at the Ice Chalet. Assume that the children in the selected class are a random sample of the population.

The 92% confidence interval is _______.

Answer

(0.72; 0.88)

8-54 §8.3 source ↗

Use the following information to answer the next 16 exercises: The Ice Chalet offers dozens of different beginning ice-skating classes. All of the class names are put into a bucket. The 5 P.M., Monday night, ages 8 to 12, beginning ice-skating class was picked. In that class were 64 girls and 16 boys. Suppose that we are interested in the true proportion of girls, ages 8 to 12, in all beginning ice-skating classes at the Ice Chalet. Assume that the children in the selected class are a random sample of the population.

In one complete sentence, explain what the interval means.

Answer

With 92% confidence, we estimate the proportion of girls, ages 8 to 12, in a beginning ice-skating class at the Ice Chalet to be between 72% and 88%.

8-56 §8.3 source ↗

Use the following information to answer the next 16 exercises: The Ice Chalet offers dozens of different beginning ice-skating classes. All of the class names are put into a bucket. The 5 P.M., Monday night, ages 8 to 12, beginning ice-skating class was picked. In that class were 64 girls and 16 boys. Suppose that we are interested in the true proportion of girls, ages 8 to 12, in all beginning ice-skating classes at the Ice Chalet. Assume that the children in the selected class are a random sample of the population.

Using the same p′ and n = 80, how would the error bound change if the confidence level were increased to 98%? Why?

Answer

The error bound would increase. Assuming all other variables are kept constant, as the confidence level increases, the area under the curve corresponding to the confidence level becomes larger, which creates a wider interval and thus a larger error.

8-58 §8.4 source ↗

Use the following information to answer the next five exercises: The standard deviation of the weights of elephants is known to be approximately 15 pounds. We wish to construct a 95% confidence interval for the mean weight of newborn elephant calves. Fifty newborn elephants are weighed. The sample mean is 244 pounds. The sample standard deviation is 11 pounds.

Identify the following:

a. $\bar{x}$ = _____

b. σ = _____

c. n = _____

Answer

a. 244

b. 15

c. 50

8-60 §8.4 source ↗

Use the following information to answer the next five exercises: The standard deviation of the weights of elephants is known to be approximately 15 pounds. We wish to construct a 95% confidence interval for the mean weight of newborn elephant calves. Fifty newborn elephants are weighed. The sample mean is 244 pounds. The sample standard deviation is 11 pounds.

Which distribution should you use for this problem?

Answer

N (244, \frac{15}{\sqrt{50}})

8-62 §8.4 source ↗

Use the following information to answer the next five exercises: The standard deviation of the weights of elephants is known to be approximately 15 pounds. We wish to construct a 95% confidence interval for the mean weight of newborn elephant calves. Fifty newborn elephants are weighed. The sample mean is 244 pounds. The sample standard deviation is 11 pounds.

What will happen to the confidence interval obtained, if 500 newborn elephants are weighed instead of 50? Why?

Answer

As the sample size increases, there will be less variability in the mean, so the interval size decreases.

8-64 §8.4 source ↗

Use the following information to answer the next seven exercises: The U.S. Census Bureau conducts a study to determine the time needed to complete the short form. The Bureau surveys 200 people. The sample mean is 8.2 minutes. There is a known standard deviation of 2.2 minutes. The population distribution is assumed to be normal.

In words, define the random variables X and $\bar{X}$ .

Answer

X is the time in minutes it takes to complete the U.S. Census short form.

\bar{X}

is the mean time it took a sample of 200 people to complete the U.S. Census short form.

8-66 §8.4 source ↗

Use the following information to answer the next seven exercises: The U.S. Census Bureau conducts a study to determine the time needed to complete the short form. The Bureau surveys 200 people. The sample mean is 8.2 minutes. There is a known standard deviation of 2.2 minutes. The population distribution is assumed to be normal.

Construct a 90% confidence interval for the population mean time to complete the forms. State the confidence interval, sketch the graph, and calculate the error bound.

Answer

CI: (7.9441, 8.4559)

This is a normal distribution curve. The peak of the curve coincides with the point 8.2 on the horizontal axis. A central region is shaded between points 7.94 and 8.46. — Figure 8.16 in the book ↗

8-68 §8.4 source ↗

Use the following information to answer the next seven exercises: The U.S. Census Bureau conducts a study to determine the time needed to complete the short form. The Bureau surveys 200 people. The sample mean is 8.2 minutes. There is a known standard deviation of 2.2 minutes. The population distribution is assumed to be normal.

If the Census did another survey, kept the error bound the same, and surveyed only 50 people instead of 200, what would happen to the level of confidence? Why?

Answer

The level of confidence would decrease because decreasing n makes the confidence interval wider, so at the same error bound, the confidence level decreases.

8-70 §8.4 source ↗

Use the following information to answer the next ten exercises: A sample of 20 heads of lettuce was selected. Assume that the population distribution of head weight is normal. The weight of each head of lettuce was then recorded. The mean weight was 2.2 pounds with a standard deviation of 0.1 pounds. The population standard deviation is known to be 0.2 pounds.

Identify the following:

a. $\bar{x}$ = ______

b. σ = ______

c. n = ______

Answer

a. $\bar{x}$ = 2.2

b. σ = 0.2

c. n = 20

8-72 §8.4 source ↗

Use the following information to answer the next ten exercises: A sample of 20 heads of lettuce was selected. Assume that the population distribution of head weight is normal. The weight of each head of lettuce was then recorded. The mean weight was 2.2 pounds with a standard deviation of 0.1 pounds. The population standard deviation is known to be 0.2 pounds.

In words, define the random variable $\bar{X}$ .

Answer

\bar{X}

is the mean weight of a sample of 20 heads of lettuce.

8-74 §8.4 source ↗

Use the following information to answer the next ten exercises: A sample of 20 heads of lettuce was selected. Assume that the population distribution of head weight is normal. The weight of each head of lettuce was then recorded. The mean weight was 2.2 pounds with a standard deviation of 0.1 pounds. The population standard deviation is known to be 0.2 pounds.

Construct a 90% confidence interval for the population mean weight of the heads of lettuce. State the confidence interval, sketch the graph, and calculate the error bound.

Answer

EBM = 0.07
CI: (2.1264, 2.2736)

This is a normal distribution curve. The peak of the curve coincides with the point 2.2 on the horizontal axis. A central region is shaded between points 2.13 and 2.27. — Figure 8.17 in the book ↗

8-76 §8.4 source ↗

Use the following information to answer the next ten exercises: A sample of 20 heads of lettuce was selected. Assume that the population distribution of head weight is normal. The weight of each head of lettuce was then recorded. The mean weight was 2.2 pounds with a standard deviation of 0.1 pounds. The population standard deviation is known to be 0.2 pounds.

In complete sentences, explain why the confidence interval in Exercise 8.74 is larger than in Exercise 8.75.

Answer

The interval is greater because the level of confidence increased. If the only change made in the analysis is a change in confidence level, then all we are doing is changing how much area is being calculated for the normal distribution. Therefore, a larger confidence level results in larger areas and larger intervals.

8-78 §8.4 source ↗

Use the following information to answer the next ten exercises: A sample of 20 heads of lettuce was selected. Assume that the population distribution of head weight is normal. The weight of each head of lettuce was then recorded. The mean weight was 2.2 pounds with a standard deviation of 0.1 pounds. The population standard deviation is known to be 0.2 pounds.

What would happen if 40 heads of lettuce were sampled instead of 20, and the error bound remained the same?

Answer

The confidence level would increase.

8-80 §8.4 source ↗

Use the following information to answer the next 14 exercises: The mean age for all Foothill College students for a recent Fall term was 33.2. The population standard deviation has been pretty consistent at 15. Suppose that twenty-five Winter students were randomly selected. The mean age for the sample was 30.4. We are interested in the true mean age for Winter Foothill College students. Let X = the age of a Winter Foothill College student.

$\bar{x}$ = _____

Answer

30.4

8-82 §8.4 source ↗

Use the following information to answer the next 14 exercises: The mean age for all Foothill College students for a recent Fall term was 33.2. The population standard deviation has been pretty consistent at 15. Suppose that twenty-five Winter students were randomly selected. The mean age for the sample was 30.4. We are interested in the true mean age for Winter Foothill College students. Let X = the age of a Winter Foothill College student.

________ = 15

Answer

σ

8-84 §8.4 source ↗

Use the following information to answer the next 14 exercises: The mean age for all Foothill College students for a recent Fall term was 33.2. The population standard deviation has been pretty consistent at 15. Suppose that twenty-five Winter students were randomly selected. The mean age for the sample was 30.4. We are interested in the true mean age for Winter Foothill College students. Let X = the age of a Winter Foothill College student.

What is $\bar{x}$ estimating?

Answer

μ

8-86 §8.4 source ↗

Use the following information to answer the next 14 exercises: The mean age for all Foothill College students for a recent Fall term was 33.2. The population standard deviation has been pretty consistent at 15. Suppose that twenty-five Winter students were randomly selected. The mean age for the sample was 30.4. We are interested in the true mean age for Winter Foothill College students. Let X = the age of a Winter Foothill College student.

As a result of your answer to Exercise 8.83, state the exact distribution to use when calculating the confidence interval.

Answer

normal

8-88 §8.4 source ↗

Construct a 95% Confidence Interval for the true mean age of Winter Foothill College students by working out then answering the next seven exercises.

How much area is in each tail? $\frac{α}{2}$ =________

Answer

0.025

8-90 §8.4 source ↗

Construct a 95% Confidence Interval for the true mean age of Winter Foothill College students by working out then answering the next seven exercises.

The 95% confidence interval is:__________________.

Answer

(24.52,36.28)

8-92 §8.4 source ↗

Construct a 95% Confidence Interval for the true mean age of Winter Foothill College students by working out then answering the next seven exercises.

In one complete sentence, explain what the interval means.

Answer

We are 95% confident that the true mean age for Winger Foothill College students is between 24.52 and 36.28.

8-94 §8.4 source ↗

Construct a 95% Confidence Interval for the true mean age of Winter Foothill College students by working out then answering the next seven exercises.

Using the same mean, standard deviation, and sample size, how would the error bound change if the confidence level were reduced to 90%? Why?

Answer

The error bound for the mean would decrease because as the CL decreases, you need less area under the normal curve (which translates into a smaller interval) to capture the true population mean.

8-96 §8.4 source ↗

Construct a 95% Confidence Interval for the true mean age of Winter Foothill College students by working out then answering the next seven exercises.

Find the value of the sample size needed to be 95% confident that the sample proportion and the population proportion are within 2% of each other. The sample proportion is 0.650. Note: Round all fractions up for n.

Answer

2,185

8-98 §8.4 source ↗

Construct a 95% Confidence Interval for the true mean age of Winter Foothill College students by working out then answering the next seven exercises.

Find the value of the sample size needed to be 90% confident that the sample proportion and the population proportion are within 1% of each other. The sample proportion is 0.50. Note: Round all fractions up for n.

Answer

6,765

8-100 §8.4 source ↗

Construct a 95% Confidence Interval for the true mean age of Winter Foothill College students by working out then answering the next seven exercises.

Find the value of the sample size needed to be 95% confident that the sample proportion and the population proportion are within 4% of each other. The sample proportion is 0.45. Note: Round all fractions up for n.

Answer

595

8-103 §8.2 source ↗

A random survey of enrollment at 35 community colleges across the United States yielded the following figures: 6,414; 1,550; 2,109; 9,350; 21,828; 4,300; 5,944; 5,722; 2,825; 2,044; 5,481; 5,200; 5,853; 2,750; 10,012; 6,357; 27,000; 9,414; 7,681; 3,200; 17,500; 9,200; 7,380; 18,314; 6,557; 13,713; 17,768; 7,493; 2,771; 2,861; 1,263; 7,285; 28,165; 5,080; 11,622. Assume the underlying population is normal.

a.

i. $\bar{x}$ = __________

ii. $s_{x}$ = __________

iii. n = __________

iv. n – 1 = __________

b. Define the random variables $X$ and $\bar{X}$ in words.

c. Which distribution should you use for this problem? Explain your choice.

d. Construct a 95% confidence interval for the population mean enrollment at community colleges in the United States.

i. State the confidence interval.

ii. Sketch the graph.

e. What will happen to the error bound and confidence interval if 500 community colleges were surveyed? Why?

Answer

a.

i. 8629

ii. 6944

iii. 35

iv. 34

b. $t_{34}$

c.

i. CI: (6244, 11,014)

ii.

d. It will become smaller

This is a normal distribution curve. The peak of the curve coincides with the point 8,629 on the horizontal axis. A central region is shaded between points 6,244 and 11,014. — Figure 8.18 in the book ↗

8-105 §8.2 source ↗

A pharmaceutical company makes tranquilizers. It is assumed that the distribution for the length of time they last is approximately normal. Researchers in a hospital used the drug on a random sample of nine patients. The effective period of the tranquilizer for each patient (in hours) was as follows: 2.7; 2.8; 3.0; 2.3; 2.3; 2.2; 2.8; 2.1; and 2.4.

a.

i. $\bar{x}$ = __________

ii. $s_{x}$ = __________

iii. n = __________

iv. n – 1 = __________

b. Define the random variable $X$ in words.

c. Define the random variable $\bar{X}$ in words.

d. Which distribution should you use for this problem? Explain your choice.

e. Construct a 95% confidence interval for the population mean length of time.

i. State the confidence interval.

ii. Sketch the graph.

f. What does it mean to be “95% confident” in this problem?

Answer

a.

i. $\bar{x}$ = 2.51

ii. $s_{x}$ = 0.318

iii. n = 9

iv. n - 1 = 8

b. the effective length of time for a tranquilizer

c. the mean effective length of time of tranquilizers from a sample of nine patients

d. We need to use a Student’s-t distribution, because we do not know the population standard deviation.

e.

i. CI: (2.27, 2.76)

ii. Answers may vary.

f. If we were to sample many groups of nine patients, 95% of the samples would contain the true population mean length of time.

8-107 §8.2 source ↗

The Federal Election Commission (FEC) collects information about campaign contributions and disbursements for candidates and political committees each election cycle. A political action committee (PAC) is a committee formed to raise money for candidates and campaigns. A Leadership PAC is a PAC formed by a federal politician (senator or representative) to raise money to help other candidates’ campaigns.

The FEC has reported financial information for 556 Leadership PACs that were operating during a certain election cycle. The following table shows the total receipts during this cycle for a random selection of 30 Leadership PACs.

$\bar{x} =$ $$ 251,$ $854.23$

$s =$ $$ 521,$ $130.41$

Use this sample data to construct a 95% confidence interval for the mean amount of money raised by all Leadership PACs during that specific election cycle. Use the Student's t-distribution.

Table 8.3 in the book ↗

$46,500.00	$0	$40,966.50	$105,887.20	$5,175.00
$29,050.00	$19,500.00	$181,557.20	$31,500.00	$149,970.80
$2,555,363.20	$12,025.00	$409,000.00	$60,521.70	$18,000.00
$61,810.20	$76,530.80	$119,459.20	$0	$63,520.00
$6,500.00	$502,578.00	$705,061.10	$708,258.90	$135,810.00
$2,000.00	$2,000.00	$0	$1,287,933.80	$219,148.30

Answer

$\bar{x} =$ $$ 251,$ $854.23$

$s =$ $$ 521,$ $130.41$

Note that we are not given the population standard deviation, only the standard deviation of the sample.

There are 30 measures in the sample, so n = 30, and df = 30 - 1 = 29

CL = 0.96, so α = 1 - CL = 1 - 0.96 = 0.04

$\frac{α}{2} =$ $0.02 t_{\frac{α}{2}} =$ $t_{0.02}$ = 2.150

$E B M =$ $t_{\frac{α}{2}} (\frac{s}{\sqrt{n}}) =$ $2.150 (\frac{521, 130.41}{\sqrt{30}}) ~ $ 204,$ $561.66$

$\bar{x}$ - EBM = $251,854.23 - $204,561.66 = $47,292.57

$\bar{x}$ + EBM = $251,854.23+ $204,561.66 = $456,415.89

We estimate with 96% confidence that the mean amount of money raised by all Leadership PACs during that specific election cycle lies between $47,292.57 and $456,415.89.

8-109 §8.2 source ↗

Unoccupied seats on flights cause airlines to lose revenue. Suppose a large airline wants to estimate its mean number of unoccupied seats per flight over the past year. To accomplish this, the records of 225 flights are randomly selected and the number of unoccupied seats is noted for each of the sampled flights. The sample mean is 11.6 seats and the sample standard deviation is 4.1 seats.

a.

i. $\bar{x}$ = __________

ii. $s_{x}$ = __________

iii. n = __________

iv. n-1 = __________

b. Define the random variables $X$ and $\bar{X}$ in words.

c. Which distribution should you use for this problem? Explain your choice.

d. Construct a 92% confidence interval for the population mean number of unoccupied seats per flight.

i. State the confidence interval.

ii. Sketch the graph.

Answer

a.

i. $\bar{x}$ = 11.6

ii. $s_{x}$ = 4.1

iii. n = 225

iv. n - 1 = 224

b. X is the number of unoccupied seats on a single flight. $\bar{X}$ is the mean number of unoccupied seats from a sample of 225 flights.

c. We will use a Student’s-t distribution, because we do not know the population standard deviation.

d.

i. CI: (11.12 , 12.08)

ii. Answers may vary.

8-111 §8.2 source ↗

Six different national brands of chocolate chip cookies were randomly selected at the supermarket. The grams of fat per serving are as follows: 8; 8; 10; 7; 9; 9. Assume the underlying distribution is approximately normal.

a. Construct a 90% confidence interval for the population mean grams of fat per serving of chocolate chip cookies sold in supermarkets.

i. State the confidence interval.

ii. Sketch the graph.

b. If you wanted a smaller error bound while keeping the same level of confidence, what should have been changed in the study before it was done?

c. Go to the store and record the grams of fat per serving of six brands of chocolate chip cookies.

d. Calculate the mean.

e. Is the mean within the interval you calculated in part a? Did you expect it to be? Why or why not?

Answer

a.

i. CI: (7.64 , 9.36)

ii.

b. The sample should have been increased.

c. Answers will vary.

d. Answers will vary.

e. Answers will vary.

This is a normal distribution curve. The peak of the curve coincides with the point 8.5 on the horizontal axis. A central region is shaded between points 7.64 and 9.36. — Figure 8.19 in the book ↗

8-113 §8.2 source ↗

Use the following information to answer the next two exercises: A quality control specialist for a restaurant chain takes a random sample of size 12 to check the amount of soda served in the 16 oz. serving size. The sample mean is 13.30 with a sample standard deviation of 1.55. Assume the underlying population is normally distributed.

Find the 95% Confidence Interval for the true population mean for the amount of soda served.

a. (12.42, 14.18)

b. (12.32, 14.29)

c. (12.50, 14.10)

d. Impossible to determine

Answer

b

8-114 §8.3 source ↗

Insurance companies are interested in knowing the population percent of drivers who always buckle up before riding in a car.

a. When designing a study to determine this population proportion, what is the minimum number you would need to survey to be 95% confident that the population proportion is estimated to within 0.03?

b. If it were later determined that it was important to be more than 95% confident and a new survey was commissioned, how would that affect the minimum number you would need to survey? Why?

Answer

a. 1,068

b. The sample size would need to be increased since the critical value increases as the confidence level increases.

8-116 §8.3 source ↗

According to a recent survey of 1,200 people, 61% feel that the president is doing an acceptable job. We are interested in the population proportion of people who feel the president is doing an acceptable job.

a. Define the random variables X and P′ in words.

b. Which distribution should you use for this problem? Explain your choice.

c. Construct a 90% confidence interval for the population proportion of people who feel the president is doing an acceptable job.

i. State the confidence interval.

ii. Sketch the graph.

Answer

a.

X = the number of people who feel that the president is doing an acceptable job;

P′ = the proportion of people in a sample who feel that the president is doing an acceptable job.

b. $N (0.61, \sqrt{\frac{(0.61) (0.39)}{1200}})$

c.

i. CI: (0.59, 0.63)

ii. Answers may vary.

8-118 §8.3 source ↗

Refer to the information in Exercise 8.117.

a. Construct three 95% confidence intervals.

i. percent of all Asians who would welcome a White person into their families.

ii. percent of all Asians who would welcome a Hispanic/Latino into their families.

iii. percent of all Asians who would welcome a Black person into their families.

b. Even though the three point estimates are different, do any of the confidence intervals overlap? Which?

c. For any intervals that do overlap, in words, what does this imply about the significance of the differences in the true proportions?

d. For any intervals that do not overlap, in words, what does this imply about the significance of the differences in the true proportions?

Answer

a.

i. (0.72, 0.82)

ii. (0.65, 0.76)

iii. (0.60, 0.72)

b. Yes, the intervals (0.72, 0.82) and (0.65, 0.76) overlap, and the intervals (0.65, 0.76) and (0.60, 0.72) overlap.

c. We can say that there does not appear to be a significant difference between the proportion of Asian adults who say that their families would welcome a White person into their families and the proportion of Asian adults who say that their families would welcome a Hispanic/Latino person into their families.

d. We can say that there is a significant difference between the proportion of Asian adults who say that their families would welcome a White person into their families and the proportion of Asian adults who say that their families would welcome a Black person into their families.

8-120 §8.3 source ↗

A telephone poll of 1,000 adult Americans was reported in an issue of Time Magazine. One of the questions asked was “What is the main problem facing the country?” Twenty percent answered “crime.” We are interested in the population proportion of adult Americans who feel that crime is the main problem.

a. Define the random variables X and P′ in words.

b. Which distribution should you use for this problem? Explain your choice.

c. Construct a 95% confidence interval for the population proportion of adult Americans who feel that crime is the main problem.

i. State the confidence interval.

ii. Sketch the graph.

d. Suppose we want to lower the sampling error. What is one way to accomplish that?

e. The sampling error given by Yankelovich Partners, Inc. (which conducted the poll) is ±3%. In one to three complete sentences, explain what the ±3% represents.

Answer

a. X = the number of adult Americans who feel that crime is the main problem; $p^{'}$ = the proportion of adult Americans who feel that crime is the main problem

b. Since we are estimating a proportion, given $p^{'}$ = 0.2 and n = 1000, the distribution we should use is $N (0.2, \sqrt{\frac{(0.2) (0.8)}{1000}})$ .

c.

i. CI: (0.18, 0.22)

ii. Answers may vary.

d. One way to lower the sampling error is to increase the sample size.

e. The stated “± 3%” represents the maximum error bound. This means that those doing the study are reporting a maximum error of 3%. Thus, they estimate the percentage of adult Americans who feel that crime is the main problem to be between 18% and 22%.

8-122 §8.3 source ↗

Use the following information to answer the next three exercises: According to a Field Poll, 79% of Georgia adults (actual results are 400 out of 506 surveyed) feel that “education and our schools” is one of the top issues facing Georgia. We wish to construct a 90% confidence interval for the true proportion of Georgia adults who feel that education and the schools is one of the top issues facing Georgia.

A point estimate for the true population proportion is:

a. 0.90

b. 1.27

c. 0.79

d. 400

Answer

c

8-125 §8.3 source ↗

Use the following information to answer the next two exercises: Five hundred and eleven (511) homes in a certain southern California community are randomly surveyed to determine if they meet minimal earthquake preparedness recommendations. One hundred seventy-three (173) of the homes surveyed met the minimum recommendations for earthquake preparedness, and 338 did not.

The point estimate for the population proportion of homes that do not meet the minimum recommendations for earthquake preparedness is ______.

a. 0.6614

b. 0.3386

c. 173

d. 338

Answer

a

8-127 §8.3 source ↗

Use the following information to answer the next two exercises: Five hundred and eleven (511) homes in a certain southern California community are randomly surveyed to determine if they meet minimal earthquake preparedness recommendations. One hundred seventy-three (173) of the homes surveyed met the minimum recommendations for earthquake preparedness, and 338 did not.

A national survey of 1,000 adults was conducted by Rasmussen Reports. It concluded with 95% confidence that 49% to 55% of Americans believe that big-time college sports programs corrupt the process of higher education.

a. Find the point estimate and the error bound for this confidence interval.

b. Can we (with 95% confidence) conclude that more than half of all American adults believe this?

c. Use the point estimate from part a and n = 1,000 to calculate a 75% confidence interval for the proportion of American adults that believe that major college sports programs corrupt higher education.

d. Can we (with 75% confidence) conclude that at least half of all American adults believe this?

Answer

a. $p^{'}$ = $\frac{(0 .55 + 0 .49)}{2}$ = 0.52; EBP = 0.55 - 0.52 = 0.03

b. No, the confidence interval includes values less than or equal to 0.50. It is possible that less than half of the population believe this.

c. CL = 0.75, so α = 1 – 0.75 = 0.25 and $\frac{α}{2} =$ $0.125 z_{\frac{α}{2}} =$ $1.150$ . (The area to the right of this z is 0.125, so the area to the left is 1 – 0.125 = 0.875.)
$E B P =$ $(1.150) \sqrt{\frac{0.52 (0.48)}{1, 000}} \approx$ $0.018$
(p′ - EBP, p′ + EBP) = (0.52 – 0.018, 0.52 + 0.018) = (0.502, 0.538)

d. Yes – this interval does not fall less than 0.50 so we can conclude that at least half of all American adults believe that major sports programs corrupt education – but we do so with only 75% confidence.

8-130 §8.4 source ↗

Among various ethnic groups, the standard deviation of heights is known to be approximately three inches. We wish to construct a 95% confidence interval for the mean height of male Swedes. Forty-eight male Swedes are surveyed. The sample mean is 71 inches. The sample standard deviation is 2.8 inches.

a.

i. $\bar{x}$ =________

ii. σ =________

iii. n =________

b. In words, define the random variables X and $\bar{X}$ .

c. Which distribution should you use for this problem? Explain your choice.

d. Construct a 95% confidence interval for the population mean height of male Swedes.

i. State the confidence interval.

ii. Sketch the graph.

e. What will happen to the level of confidence obtained if 1,000 male Swedes are surveyed instead of 48? Why?

Answer

a.

i. 71

ii. 2.8

iii. 48

b. X is the height of a male Swede, and $\overline{x}$ is the mean height from a sample of 48 male Swedes.

c. Normal. We know the standard deviation for the population, and the sample size is greater than 30.

d.

i. CI: (70.151, 71.85)

ii.

e. The confidence interval will decrease in size, because the sample size increased. Recall, when all factors remain unchanged, an increase in sample size decreases variability. Thus, we do not need as large an interval to capture the true population mean.

This is a normal distribution curve. A central region is shaded between points 70.15 and 71.85. — Figure 8.20 in the book ↗

8-132 §8.4 source ↗

Suppose that an accounting firm does a study to determine the time needed to complete one person’s tax forms. It randomly surveys 100 people. The sample mean is 23.6 hours. There is a known standard deviation of 7.0 hours. The population distribution is assumed to be normal.

a.

i. $\bar{x}$ =________

ii. σ =________

iii. n =________

b. In words, define the random variables X and $\bar{X}$ .

c. Which distribution should you use for this problem? Explain your choice.

d. Construct a 90% confidence interval for the population mean time to complete the tax forms.

i. State the confidence interval.

ii. Sketch the graph.

e. If the firm wished to increase its level of confidence and keep the error bound the same by taking another survey, what changes should it make?

f. If the firm did another survey, kept the error bound the same, and only surveyed 49 people, what would happen to the level of confidence? Why?

g. Suppose that the firm decided that it needed to be at least 96% confident of the population mean length of time to within one hour. How would the number of people the firm surveys change? Why?

Answer

a.

i. $\bar{x}$ = 23.6

ii. $σ$ = 7

iii. n = 100

b. X is the time needed to complete an individual tax form. $\bar{X}$ is the mean time to complete tax forms from a sample of 100 customers.

c. $N (23.6, \frac{7}{\sqrt{100}})$ because we know sigma.

d.

i. (22.228, 24.972)

ii.

e. It will need to change the sample size. The firm needs to determine what the confidence level should be, then apply the error bound formula to determine the necessary sample size.

f. The confidence level would increase as a result of a larger interval. Smaller sample sizes result in more variability. To capture the true population mean, we need to have a larger interval.

g. According to the error bound formula, the firm needs to survey 206 people. Since we increase the confidence level, we need to increase either our error bound or the sample size.

This is a normal distribution curve. A central region is shaded between points 22.228 and 24.972. — Figure 8.21 in the book ↗

8-134 §8.4 source ↗

A camp director is interested in the mean number of letters each child sends during their camp session. The population standard deviation is known to be 2.5. A survey of 20 campers is taken. The mean from the sample is 7.9 with a sample standard deviation of 2.8.

a.

i. $\bar{x}$ =________

ii. σ =________

iii. n =________

b. Define the random variables X and $\bar{X}$ in words.

c. Which distribution should you use for this problem? Explain your choice.

d. Construct a 90% confidence interval for the population mean number of letters campers send home.

i. State the confidence interval.

ii. Sketch the graph.

e. What will happen to the error bound and confidence interval if 500 campers are surveyed? Why?

Answer

a.

i. 7.9

ii. 2.5

iii. 20

b. X is the number of letters a single camper will send home. $\bar{X}$ is the mean number of letters sent home from a sample of 20 campers.

c.

N $7.9 (\frac{2.5}{\sqrt{20}})$

d.

i. CI: (6.98, 8.82)

ii.

e. The error bound and confidence interval will decrease.

This is a normal distribution curve. A central region is shaded between points 6.98 and 8.82. — Figure 8.22 in the book ↗

8-136 §8.4 source ↗

The Federal Election Commission collects information about campaign contributions and disbursements for candidates and political committees each election cycle. During a certain campaign season, there were 1,619 candidates for the House of Representatives across the United States who received contributions from individuals. Table 8.5 shows the total receipts from individuals for a random selection of 40 House candidates rounded to the nearest $100. The standard deviation for this data to the nearest hundred is σ = $909,200.

a. Find the point estimate for the population mean.

b. Using 95% confidence, calculate the error bound.

c. Create a 95% confidence interval for the mean total individual contributions.

d. Interpret the confidence interval in the context of the problem.

Table 8.5 in the book ↗

$3,600	$1,243,900	$10,900	$385,200	$581,500
$7,400	$2,900	$400	$3,714,500	$632,500
$391,000	$467,400	$56,800	$5,800	$405,200
$733,200	$8,000	$468,700	$75,200	$41,000
$13,300	$9,500	$953,800	$1,113,500	$1,109,300
$353,900	$986,100	$88,600	$378,200	$13,200
$3,800	$745,100	$5,800	$3,072,100	$1,626,700
$512,900	$2,309,200	$6,600	$202,400	$15,800

Answer

a. $\bar{x}$ = $568,873

b. CL = 0.95 α = 1 – 0.95 = 0.05 $z_{\frac{α}{2}}$ = 1.96
EBM = $z_{0.025} \frac{σ}{\sqrt{n}}$ = 1.96 $\frac{909200}{\sqrt{40}}$ = $281,764

c. $\bar{x}$ − EBM = 568,873 − 281,764 = 287,109
$\bar{x}$ + EBM = 568,873 + 281,764 = 850,637

d. We estimate with 95% confidence that the mean amount of contributions received from all individuals by House candidates is between $287,109 and $850,637.

8-139 §8.4 source ↗

If the confidence interval is change to a higher probability, would this cause a lower, or a higher, minimum sample size?

Answer

Higher

8-140 §8.4 source ↗

If the tolerance is reduced by half, how would this affect the minimum sample size?

Answer

It would increase to four times the prior value.

8-141 §8.4 source ↗

If the value of p is reduced, would this necessarily reduce the sample size needed?

Answer

No, It could have no affect if it were to change to 1 – p, for example. If it gets closer to 0.5 the minimum sample size would increase.

8-142 §8.4 source ↗

Is it acceptable to use a higher sample size than the one calculated by

\frac{z^{2} p q}{e^{2}}

?

Answer

Yes

8-143 §8.4 source ↗

A company has been running an assembly line with 97.42%% of the products made being acceptable. Then, a critical piece broke down. After the repairs the decision was made to see if the number of defective products made was still close enough to the long standing production quality. Samples of 500 pieces were selected at random, and the defective rate was found to be 0.025%.

a. Is this sample size adequate to claim the company is checking within the 90% confidence interval?

b. The 95% confidence interval?

Answer

a. No

b. No